﻿ SQL: Ascending by job name and descending by employee ID

# SQL Exercise: Ascending by job name and descending by employee ID

## SQL employee Database: Exercise-75 with Solution

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75. From the following table, write a SQL query to list the employees in the ascending order by job title and in descending order by employee ID. Return complete information about the employees.

Sample table: employees

Pictorial Presentation:

Sample Solution:

``````SELECT *
FROM employees e
ORDER BY e.job_name ASC,
e.emp_id DESC ;
``````

Sample Output:

``` emp_id | emp_name | job_name  | manager_id | hire_date  | salary  | commission | dep_id
--------+----------+-----------+------------+------------+---------+------------+--------
69062 | FRANK    | ANALYST   |      65646 | 1991-12-03 | 3100.00 |            |   2001
67858 | SCARLET  | ANALYST   |      65646 | 1997-04-19 | 3100.00 |            |   2001
69324 | MARKER   | CLERK     |      67832 | 1992-01-23 | 1400.00 |            |   1001
69000 | JULIUS   | CLERK     |      66928 | 1991-12-03 | 1050.00 |            |   3001
68736 | ADNRES   | CLERK     |      67858 | 1997-05-23 | 1200.00 |            |   2001
63679 | SANDRINE | CLERK     |      69062 | 1990-12-18 |  900.00 |            |   2001
67832 | CLARE    | MANAGER   |      68319 | 1991-06-09 | 2550.00 |            |   1001
66928 | BLAZE    | MANAGER   |      68319 | 1991-05-01 | 2750.00 |            |   3001
65646 | JONAS    | MANAGER   |      68319 | 1991-04-02 | 2957.00 |            |   2001
68319 | KAYLING  | PRESIDENT |            | 1991-11-18 | 6000.00 |            |   1001
68454 | TUCKER   | SALESMAN  |      66928 | 1991-09-08 | 1600.00 |       0.00 |   3001
66564 | MADDEN   | SALESMAN  |      66928 | 1991-09-28 | 1350.00 |    1500.00 |   3001
65271 | WADE     | SALESMAN  |      66928 | 1991-02-22 | 1350.00 |     600.00 |   3001
64989 | ADELYN   | SALESMAN  |      66928 | 1991-02-20 | 1700.00 |     400.00 |   3001
(14 rows)
```

Explanation:

The provided query in SQL that selects all columns from the 'employees' table and order the results first in ascending order on "job_name" column then for the group of same job_name the records will be sorted by the "emp_id" column in descending order.

It can be said that the records with the same "job_name" value makes a groups together, and the groups appears in alphabetical order and within each group of records with the same "job_name" value the the "emp_id" column will come from highest to lowest value.

Relational Algebra Expression:

Relational Algebra Tree:

## Practice Online

Sample Database: employee

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Previous SQL Exercise: Sort the employees in ascending order on the salary.
Next SQL Exercise: Sort unique jobs by department in descending order.

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## SQL: Tips of the Day

Grouped LIMIT in PostgreSQL: Show the first N rows for each group?

```db=# SELECT * FROM xxx;
id | section_id | name
----+------------+------
1 |          1 | A
2 |          1 | B
3 |          1 | C
4 |          1 | D
5 |          2 | E
6 |          2 | F
7 |          3 | G
8 |          2 | H
(8 rows)
```

I need the first 2 rows (ordered by name) for each section_id, i.e. a result similar to:

```id | section_id | name
----+------------+------
1 |          1 | A
2 |          1 | B
5 |          2 | E
6 |          2 | F
7 |          3 | G
(5 rows)
```

PostgreSQL v9.3 you can do a lateral join

```select distinct t_outer.section_id, t_top.id, t_top.name from t t_outer
join lateral (
select * from t t_inner
where t_inner.section_id = t_outer.section_id
order by t_inner.name
limit 2
) t_top on true
order by t_outer.section_id;
```

Database: PostgreSQL

Ref: https://bit.ly/3AfYwZI

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